Question

A student has 8.85 g of zinc powder, 1.95 L of a 2.00 M

calcium nitrate solution, and 1.35 L of a 3.10 M lead(II) nitrate solution.

Write the chemical reaction that occurs between the two reacting species. Include physical states.

What volume of the solution is needed to completely react all 8.85 g of zinc powder?

Calculate the molarity of the Zn2+ ions in solution after the reaction.

Answer 1

Zinc powder only reacts with lead(II) nitrate according to the equation:

Zn _(solid) + Pn(NO₃)_{2 (solution)} = Zn(NO₃)_{2 (solution)} + Pb _(solid)

$n(Zn)=\dfrac{Mass}{Atomic\ mass}=\dfrac{8.85\ g}{65\ g/mol}=0.1362\ mol$

The amount of Pb(NO₃)₂ needed to completely react with zinc powder:

$n(Pb(NO_3)_2)=n(Zn)=0.1362\ mol$

The volume of the solution:

$V(Pb(NO_3)_2)=\dfrac{Amount}{Concentration}=\dfrac{0.1362\ mol}{3.10\ mol/L}=0.044\ L$

When 1.35 L of a 3.10 M lead(II) nitrate solution reacted with zinc powder, Zn is the limiting reactant. The molarity of the Zn²⁺ ions in solution after the reaction:

$c(Zn^{2+})=\dfrac{Amount}{Volume}=\dfrac{0.1362\ mol}{1.35\ L}=0.1\ M$